Statistik 1 10 12 edited_anova

12 - 1
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.

12 - 2
When you have completed this chapter, you will be able to:
1. Discuss the general idea of analysis of variance.
2. List the characteristics of the F distribution.
3. Conduct a test of hypothesis to determine whether the
variances of two populations are equal.
Organize data into a one-way and a
two-way ANOVA table.
4.

12 - 3
5. Define the terms treatments and blocks.
6. Conduct a test of hypothesis to determine whether
three or more treatment means are equal.
7. Develop multiple tests for difference between each pair
of treatment means.

12 - 4
Karakteristik
Karakteristik
Distribusi-F
Distribusi-F
Terdapat Terdapat ‘‘kkeelluuaarrggaa’’ FF--DDiissttrriibbuuttiioonnss::
Each member of the family is determined by
two parameters:
…the numerator degrees of freedom, and the
… denominator degrees of freedom
F cannot be negative, and it is a continuous distribution
The F distribution is positively skewed
Its values range from 0 to ¥ as F ® ¥, the
curve approaches the X-axis

12 - 5
The F-Distribution, F(m,n)
0 1.0
Not symmetric
(skewed to the right)
Each member of the family is
determined by two parameters: the
numerator degrees of freedom (m)
and the denominator degrees of
freedom (n).
F
Nonnegative value s only
a

12 - 6
TTeesstt ffoorr EEqquuaall VVaarriiaanncceess
For the two tailed test, the test statistic
F =s
2 and are the sample variances for the two samples
s1 2
is given by:
2
1
s
2
2
The null hypothesis is rejected
if the computed value of the test statistic
is greater than the critical value
s2

12 - 7
Test for Equal Variances
• For the two tail test, the test statistic is given by:
• where s1
2 and s2
arg ( , )
L er of S S
= 2
2
1
s
2 are the sample variances for the two samples.
• The null hypothesis is rejected at a level of significance if the
computed value of the test statistic is greater than the critical
value with a confidence level a/2 and numerator and
denominator dfs.
( , )
2
2
1
2
2
2
1
Smaller of S S
F
2
2
F =s

12 - 8
Test for Equal Variances
• For the one tail test, the test statistic is given by:
where s1
2 and s2
2
S
= 1 if H : σ > σ
S
2 are the sample variances for the two samples.
• The null hypothesis is rejected at a level of significance if the
computed value of the test statistic is greater than the critical
value with a confidence level a and numerator and
denominator dfs.
2
2
2
2 1 1
2
F

12 - 9
Bejo, seorang pialang di Maju Securities,
mencatat rata-rata return pada sebuah sample
10 saham internet 12.6 persen dengan standar
deviasi 3.9 persen.
Sedangkan rata-rata return pada sebuah sample
8 saham utilitas adalah 10.9 persen dengan
standar deviasi 3.5 persen.
Pada tingkat signifikansi 0.05,
dapatkah Bejo menyimpulkan bahwa terdapat
variasi yang lebih besar pada saham internet?

12 - 10
HHyyppootthheessiiss TTeessttiinngg
SStteepp 11 SSttaattee tthhee nnuullll aanndd aalltteerrnnaattee hhyyppootthheesseess
SStteepp 22 SSeelleecctt tthhee lleevveell ooff ssiiggnniiffiiccaannccee
SStteepp 33 IIddeennttiiffyy tthhee tteesstt ssttaattiissttiicc
SStteepp 44 SSttaattee tthhee ddeecciissiioonn rruullee
SStteepp 55
Compute the value of the test statistic
Compute the value of the test statistic
and make a decision
and make a decision
Do not reject H0 Do not reject H0 Reject H0 and accept H1 Reject H0 and accept H1

12 - 11
H Hyyppootthheessisis T Teesstt
SStteepp 11
State State the the null null and and alternate
alternate
hypotheses
hypotheses
Compute the test
statistic and make
a decision
2
1
s
H 0: s I £ s U
H1 : I s > s
= s =
2 2
2 2
2
(3.9)
U
F == 11..22441166 2
a = 0.05
The test statistic is the
F distribution
Compute the test
statistic and make
a decision
SStteepp 55
Reject H0 if F > 3.68 The df
are 9 in the numerator and
7 in the denominator.
2
2
(3.5)
Do not reject the null hypothesis; there is insufficient
evidence to show more variation in the internet stocks.

12 - 12
Contoh
NNiillaaii UUjjiiaann
KKeellaass AA KKeellaass BB
5522 5599
6677 6600
5566 6611
4455 5511
7700 5566
5544 6633
6644 5577
6655
Uji apakah ada perbedaan yang
signifikan varians nilai ujian
kelas A dan kelas B ?

12 - 13
The F distribution is also used for testing
whether
The F distribution is also used for testing
whether
two or more sample means
two or more sample means
came from
came from
the same or equal
populations
This this technique is called
analysis of variance or ANOVA
the same or equal
populations
ANOVA

12 - 14
ANOVA
ANOVA
requires the following
requires the following
conditions…
conditions…
…the sampled populations follow the
normal distribution
…the populations have equal standard deviations
…the samples are randomly selected
and are independent

12 - 15
• ANOVA requires the following conditions:
– The populations being sampled are normally
distributed.
– The populations have equal standard deviations.
– The samples are randomly selected and are
independent
– Data must be at least interval-scale.
• Type of ANOVA :
– One-Way (One-Factor) ANOVA
– Multi-Way (Multi-Factor) ANOVA
 Two-Way ANOVA

12 - 16
One Way ANOVA
TTRREEAATTMMEENNTT
11 22 33
XX11..11 XX11..22 XX11..33
XX22..11 XX22..22 XX11..11
XX33..11 XX33..22 XX11..11
XX44..11 XX44..22 XX11..11
TT11 TT22 TTj 33 T
j X 1 X 2 X 3 X
X : Overall Mean (Grand Mean);
T
X
T =SX

12 - 17
AANNOOVVAA PPrroocceedduurree
The Null Hypothesis (H0) is that the population
means are the same
The Alternative Hypothesis (H1) is that
at least one of the means is different
The Test Statistic is the F distribution
The Decision rule is to reject H0
if
F(computed) is greater than F(table)
with numerator and denominator df

12 - 18
The hypothesis
• Suppose that we have independent samples of n1,
n2, . . ., nK observations from K populations. If the
population means are denoted by m1, m2, . . ., mK, the
one-way analysis of variance framework is
designed to test the null hypothesis
: = = =
H μ μ μ
0 1 2
: For at least one pair ,
H μ μ μ μ
1 i j
i j
K
≠


12 - 19
Sample Observations from Independent
Random Samples of K Populations
Population 1 2 . . . K
Mean m1 m2 . . . mK
Variance s2 s2 . . . s2
Sample
observations
from the
population
x11
x12
.
.
.
x1n1
x21
x22
.
.
.
x2n2
. . .
. . .
. . .
xK1
xK2
.
.
.
xKnK
Sample size n1 n2 . . . nK
Same !!
unequal !!
Unequal number of observations in the K samples in general.
nT=n1+…+nK

12 - 20

12 - 21
Terminology
Total Variation …is the sum of the squared differences
between each observation and
Treatment
Variation …is the sum of the squared differences
between each treatment mean and
the overall mean
the overall mean
Random Variation …is the sum of the squared differences
between each observation and
its treatment mean

12 - 22
 If there k populations being sampled, the
numerator df = k – 1
 If there are a total of n observations the
denominator df =n - k
 The test statistic is computed by:
= (k-1)
F SSTSSE
(n- k)

12 - 23
• SS Total is the total sum of squares
S
X X
n
2
SSTotal = S 2 -
( )

12 - 24
• SST is the treatment sum of squares
SST T
( )
2 S 2 - ÷ ÷ø
n X
c
n
c
ö
ç çè= S æ
TC is the column total, nc is the number of
observations in each column,
SX the sum of all the observations, and
n the total number of
observations

12 - 25
•SSE is the sum of squares error
SSE = SS total - SST

12 - 26
One-Way ANOVA Table
SSoouurrccee ooff
VVaarriiaattiioonn
SSuumm ooff
SSqquuaarree
dd..ff.. MMeeaann SSqquuaarree FF
TTrreeaattmmeenntt SSSSTT kk –– 11 MMSSTT == SSSSTT//((kk –– 11)) MMSSTT//MMSSEE
EErrrroorr SSSSEE nnTT -- kk MMSSEE == SSSSEE//((NN –– kk))
TToottaall SSSSttoottaall nT - 1

12 - 27
D’Cost memiliki spesialisasi makanan murah meriah.
Bejo, Manager D’cost, baru saja membuat menu Nasi Goreng
baru. Sebelum dimasukkan dalam menu reguler, dia
memutuskan mencoba menawarkannya di beberapa
restorannya.
Bejo ingin mengetahui apakah ada perbedaan jumlah rata-rata nasi
goreng baru terjual per hari di restoran D’cost Kaza, Royal, dan
Kayon restaurants.
Gunakan tingkat signifikansi 5%.

12 - 28
Kaza Royal Kayon
13 10 18
12 12 16
14 13 17
12 11 17
Tc 51 46 85
nc 4 4 5
17
Tc 51 46 85
nc 4 4 5
…continued

12 - 29
KAZA ROYAL KAYON KAZA^2 ROYAL^2 KAYON^2
13 10 18 169 100 324
12 12 16 144 144 256
14 13 17 196 169 289
12 11 17 144 121 289
17 289
Tc 51 46 85 182 653 534 1447 2634
nc 4 4 5 13

12 - 30
…continued
• SS Total (is the total sum of squares)
SS Total = S X 2 - ( S X
) 2 n
= 86
13
= 2634 -
(182)2

12 - 31
…continued
•SST is the treatment sum of squares
= S æ
SST T
ö
51 2 46
2 2 2
( )
2 S 2 - ÷ ÷ø
n X
c
n
c
ö
ç çè
( ) ( ) ( )
= 76.25
(182 )
13
85
5
4
4
- ÷ ÷ø
ç çè æ
= + +

12 - 32
…continued
•SSE is the sum of squares error
SSE = SS Total - SST
86 – 76.25
= 9.75

12 - 33
SStteepp 11
alternate
hypotheses
hypotheses
Compute the test
statistic and make
a decision
H0: 1 m 2 =m = 3 m
a = 0.05
F distribution
Compute the test
statistic and make
a decision
SStteepp 55
Reject H0 if F > 4.10 The df
are 2 in the numerator and
10 in the denominator.
== 3399..1100
H1 :
Treatment means are
not all equal
F SST
SSE
( n- k )
=
(k - 1)
= 76.25 2
9.75 10

12 - 34
…continued
 The decision is to reject the null hypothesis
The treatment means are not the same
 The mean number of meals sold at the three
locations is not the same

12 - 35
AANNOOVVAA TTaabbllee
…from the Minitab system
Analysis of Variance
Source DF SS MS F P
Factor 2 76.250 38.125 39.10 0.000
Error 10 9.750 0.975
Total 12 86.000
Individual 95% CIs For Mean Based on Pooled St.Dev
Level N Mean St.Dev ---------+---------
+---------+-------
Kaza 4 12.750 0.957 (---*---)
Royal 4 11.500 1.291 (---*---)
Kayon 5 17.000 0.707 (---*---)
---------+---------+---------+-------
Pooled St.Dev = 0.987 12.5 15.0 17.5

12 - 36
One-Way ANOVA Table
SSoouurrccee ooff
VVaarriiaattiioonn
SSuumm ooff
SSqquuaarree
dd..ff.. MMeeaann SSqquuaarree FF
TTrreeaattmmeenntt 7766..2255 22 MMSSTT == 3388..1122 3399..0099
EErrrroorr 99..7755 1100 MMSSEE == 00..997755
TToottaall 8866..0000 12

12 - 37
Inferences
About
Treatment
Means

12 - 38
Inferences
About
Treatment
Means
Ketika menolak hipothesis null bahwa rata-ratanya
sama, kita mungkin juga ingin
mengetahui rata-rata treatment mana yang
berbeda
Salah satu prosedur yang paling
sederhana adalah melalui penggunaan
confidence intervals
CCoonnffiiddeennccee IInntteerrvvaall

12 - 39
Confidence Interval for the
Difference Between Two Means
(X1 - X2)± t MSE n n 1 2
where t is obtained from the t table
with degrees of freedom (n - k).
where t is obtained from the t table
with degrees of freedom (n - k).
MSE = [SSE/(n - k)]
MSE = [SSE/(n - k)]
1 1 + æè ç
öø ÷

12 - 40
Develop a
95% confidence interval
for the
difference in the mean number
of Nasi Goreng sold in Kayon and
Kaza.
Can Bejo conclude that there is a
difference between the two restaurants?

12 - 41
(X X ) 1 2 - ± t MSE
MMSSEE
1 1
+
n1 n2
æ
è ç
ö
ø ÷
(17-12.75) ± 2.228 .975
1
4
1
5
+
æ
è ç
ö
ø ÷
4 .25 ± 1.48 Þ ( 2.77, 5.73)

12 - 42
Contoh
• Berikut adalah hsl panen padi (kuintal) dari 12 petak
sawah dengan 3 jenis pupuk yang berbeda. Tiap jenis
pupuk diberikan pada masing-masing 4 petak sawah.
Apakah ada perbedaan hsl panen dari ketiga jenis pupuk
tsb? Gunakan a = 5%
JJeenniiss PPuuppuukk
AA BB CC
5555 6666 4477
5544 7766 5511
5599 6677 4466
5566 7711 4488

12 - 43
Latihan Soal
• Berikut adalah waktu (menit) yang dibutuhkan untuk
mengerjakan 1 soal latihan dari 4 mata kuliah. Sampel
random masing-masing 5 mhs untuk tiap mata kuliah. Uji
apakah ada perbedaan yang signifikan ? (Kerjakan
menggunakan Tabel ANOVA)
EE.. MMaakkrroo EE..MMiikkrroo MMaatteemmaattiikkaa SSttaattiissttiikkaa
1188 2200 2200 2222
2211 2222 2244 2244
2200 2233 2255 2233
2255 2211 2288 2255
2266 2244 2288 2255

12 - 44
Two-Way ANOVA
TTRREEAATTMMEENNTT
11 22 33 TTii
BB
LL
OO
CC
KK
11 XX11..11 XX11..22 XX11..33 TT11
22 XX22..11 XX22..22 XX11..11 TT22
33 XXXXXXTT33..11 33..22 11..11 33
44 XXXXXXTTTTX
44..11 44..22 11..11 44
j TTTTTTjj 11 22 33
i X
T
1 X
2 X
3 X
4 X
1 X X 3 X 2 X

12 - 45
TTwwoo FFaaccttoorr AANNOOVVAA
Untuk ANOVA dua-faktor kita menguji:
apakah ada perbedaan yang signifikan pada treatment
effect dan
apakah ada perbedaan pada blocking effect.
SSB
B
n
…Let Br be the block totals (r for rows),
nr be the number of observations in each row
…Let SSB represent the sum of squares for the blocks
SST k
Xn
= éë
ê
r ùû
S ú- S 2 ( )2
( -
1
)
= (( - 1)( -
1))
SSE k b
F
r

12 - 46
General Format of Two-Way Analysis of
Variance Table
Source of
Variation
Sums of
Squares
Degrees
of
Freedom
Mean Squares F Ratios
Treatment
s
SST k-1 MST=SST/(k-1) MST/MSE
Blocks SSB b-1 MSB=SSB/(b-1) MSB/MSE
Error SSE (k-1)(b-1) MSE=SSE/[(k-1)(b-
1)]
Total SSTotal nT-1
k : number of treatment (column)
b : number of block (row)
n : number of observation

12 - 47
TTwwoo FFaaccttoorr AANNOOVVAA
Pabrik Tempe Bieber beroperasi 24 jam sehari
selama 5 hari dalam seminggu. Para pekerja
bergantian shifts kerja tiap minggu.
Pak Todd Bieber, pemilik pabrik, ingin
mengetahui apakah ada perbedaan dalam jumlah
produksi Tempe ketika para pegawai bekerja pada
shift yang berbeda.
Sampel yang terdiri dari lima pegawai dipilih dan
produksi mereka dicatat pada setiap shift. Pada
tingkat signifikansi 0.05, apakah kita dapat
menyimpulkan bahwa ada perbedaan rata-rata
produksi pada shift dan rata-rata produksi pada
pegawai?

12 - 48
…continued
AANNOOVVAA
Employee Day
Output
Evening
Output
Night
Output
McCartney 31 25 35
Neary 33 26 33
Schoen 28 24 30
Thompson 30 29 28
Wagner 28 26 27

12 - 49
Day Evening Night Tr nr Day^2 Eve^2 Night^2
McCartney 31 25 35 91 3
961
625
1,225
Neary 33 26 33 92 3
1,089
676
1,089
Schoen 28 24 30 82 3
784
576
900
Thompson 30 29 28 87 3
900
841
784
wagner 28 26 27 81 3
784
676
729
Tc 150 130 153 433
4,518
3,394
4,727
12,639
nc 5 5 5 15

12 - 50
– Compute the various sum of squares:
Source SS df MS F p-value
Treatments 62.53 2 31.267 5.75 .0283
Blocks 33.73 4 8.433 1.55 .2762
Error 43.47 8 5.433
Total 139.73 14 　　　

12 - 51
DDiiffffeerreennccee bbeettwweeeenn vvaarriioouuss sshhiiffttss??
SStteepp 11
alternate
hypotheses
hypotheses
Compute the test
statistic and make
a decision
H0: 1 m 2 =m = 3 m
a = 0.05
F distribution
Compute the test
statistic and make
a decision
SStteepp 55
Reject H0 if F > 4.46.
The df are 2
and 8
H1 :
Not all means are equal
SST ( k
-
1
)
= (( - 1)( -
1))
SSE k b
F

12 - 52
SST ( k
-
1
)
= (( - 1)( -
1))
SSE k b
-
62 .53 3 1
Since 5.754 > 4.46, H0 is rejected.
( )
43.47 ((3 - 1)(5 -
1))
=
AANNOOVVAA
…continued
SStteepp 55
There is a difference in the mean number of units
produced on the different shifts.
F
= 5.754

12 - 53
DDiiffffeerreennccee bbeettwweeeenn vvaarriioouuss sshhiiffttss??
SStteepp 11
alternate
hypotheses
hypotheses
Compute the test
statistic and make
a decision
H0: 1 m 2 =m = 3 m
a = 0.05
F distribution
Compute the test
statistic and make
a decision
SStteepp 55
H1 :
Not all means are equal
SST ( k
-
1
)
= (( - 1)( -
1))
SSE k b
F
Reject H0 if F > 3.84
The df are 4 and 8

12 - 54
SST k
= 33.73 4
43.47 (2)(4) = 1.55
AANNOOVVAA
…continued
SStteepp 55
( -
1
)
= (( - 1)( -
1))
SSE k b
F
Since 1.55 < 3.84, H0 is not rejected.
There is no significant difference in the mean
number of units produced by the various employees.

12 - 55
…from the Minitab system
Units versus Worker, Shift
Units versus Worker, Shift
Analysis of Variance for Units
Source DF SS MS F P
Worker 4 33.73 8.43 1.55 0.276
Shift 2 62.53 31.27 5.75 0.028
Error 8 43.47 5.43
Total 14 139.73
Analysis of Variance for Units
Source DF SS MS F P
Worker 4 33.73 8.43 1.55 0.276
Shift 2 62.53 31.27 5.75 0.028
Error 8 43.47 5.43
Total 14 139.73
AANNOOVVAA

12 - 56
Using
Highlight ANOVA: TWO FACTOR WITHOUT REPLICATION
SSeeee……
…Click OK
SSeelleecctt
IINNPPUUTT D DAATTAA

12 - 57
SSB
SST
SSE
SSSS TToottaall
Using
Since F(test) < F(critical),
there is not sufficient
evidence to reject H0
There is no significant
difference in the average
number of units produced
by the different employees.
FFtteesstt FFccrriittiiccaall
Since F(test) < F(critical),
there is not sufficient
evidence to reject H0
There is no significant
difference in the average
number of units produced
by the different employees.

12 - 58
Latihan soal
• Berikut adalah waktu (menit) yang dibutuhkan untuk
mengerjakan 1 soal latihan dari 4 mata kuliah. Sampel random
masing-masing 5 mhs untuk tiap mata kuliah. Uji apakah ada
perbedaan yang signifikan waktu pengerjaan antara keempat
mata kuliah dan antara mahasiswa tersebut?
EE.. MMaakkrroo EE..MMiikkrroo MMaatteemmaattiikkaa SSttaattiissttiikkaa
AA 1188 2200 2200 2222
BB 2211 2222 2244 2244
CC 2200 2233 2255 2233
DD 2255 2211 2288 2255
EE 2266 2244 2288 2255

12 - 59
Latihan Soal
• Uji dengan a = 0,1 apakah ada perbedaan hsl produksi
(kg) antara ketiga mesin dan kelima karyawan
tersebut?
MMeessiinn
KKaarryyaawwaann II IIII IIIIII
AA 2211 1177 3311
BB 2277 2255 2288
CC 2299 2200 3322
DD 2233 1155 3300
EE 2255 2233 2244

12 - 60
TTeesstt yyoouurr lleeaarrnniinngg……
Click on… Click on…
www.mcgrawhill.ca/college/lind
Online Learning Centre
for quizzes
extra content
data sets
searchable glossary
access to Statistics Canada’s E-Stat data
…and much more!

12 - 61
This completes Chapter 12

Statistik 1 10 12 edited_anova

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (19)

Similar to Statistik 1 10 12 edited_anova

Similar to Statistik 1 10 12 edited_anova (20)

Recently uploaded

Recently uploaded (20)

Statistik 1 10 12 edited_anova